On a theorem of Redheffer concerning diagonal stability

Linear Algebra and its Applications 431 (2009) 2317–2329

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Linear Algebra and its Applications journal homepage: www.elsevier.com/locate/laa

On a theorem of Redheffer concerning diagonal stability Robert Shorten a,∗ , Kumpati S. Narendra b a b

The Hamilton Institute, NUI Maynooth, Ireland Center for Systems Science, Yale University, USA

A R T I C L E

I N F O

Article history: Received 17 October 2008 Accepted 25 February 2009 Available online 26 April 2009 Submitted by Avi Berman Dedicated to Professor Shmuel Friedland on the occasion of his 65th birthday Keywords: Diagonal stability KYP Lemma

A B S T R A C T

An important problem in system theory concerns determining whether or not a given LTI system x˙ = Ax is diagonally stable. More precisely, this problem is concerned with determining conditions on a matrix A such that there exists a diagonal matrix D with positive diagonal entries (i.e. a positive diagonal matrix), satisfying AT D + DA = −Q < 0, Q = Q T > 0. While this problem has attracted much attention over the past half century, two results of note stand out: (i) a result based on Theorems of the Alternative derived by Barker et al.; and (ii) algebraic conditions derived by Redheffer. This paper is concerned with the second of these conditions. Our principal contribution is to show that Redheffer’s result can be obtained from the Kalman–Yacubovich–Popov lemma. We then show that this method of proof leads to natural generalisations of Redheffer’s result and we use these results to derive new conditions for diagonal and Hurwitz stability for special classes of matrices. © 2009 Elsevier Inc. All rights reserved.

1. Introduction It is well known that the differential equation x˙ = Ax, where A ∈ Rn×n has eigenvalues in the open left half of the complex plane, is Hurwitz stable with Lyapunov functions of the form V (x) = xT Px, where the real positive definite symmetric matrix P satisfies, Q = Q T > 0. AT P

+ PA = −Q < 0.

(1)

A question that has attracted a great deal of attention in the past [1–6], and continues to do so [7,8] in diverse application areas, including biology [7], communication networks [9], economics [10], and control engineering [11,12], concerns what additional conditions on A are required so that there exists a real positive diagonal matrix D = P satisfying (1). If such a matrix exists, the system is said to be ∗ Corresponding author. E-mail address: [email protected] (R. Shorten). 0024-3795/$ - see front matter © 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.laa.2009.02.035

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diagonally stable, and (by an abuse of the language) the Lyapunov function is called a diagonal Lyapunov function. Previous work on the diagonal stability problem has followed two distinct lines of enquiry. First, following the publication of [4], a number of authors attempted to derive algebraic conditions to verify the main result of this paper; namely that the matrix AX should have at least one negative diagonal entry for all positive semi-definite X ∈ Rn×n . Results obtained in this direction include the work of Kraaijvanger [3], Wanat [8] and others. An alternative approach was described by Redheffer in [1]. Motivated by the work of Cross, Redheffer made use of both A and A−1 in deriving conditions for diagonal stability. In particular, conditions for diagonal stability are given in his paper in terms of a CQLF existence problem for a pair of lower dimensional systems constructed from A and A−1 . The main contribution in this paper shows that Redheffer’s result also follows from basic results from systems theory. Then it is shown that this method of proof leads to natural generalisations of Redheffer’s result, and to conditions for diagonal and Hurwitz stability for special classes of matrices. 2. Notation and mathematical preliminaries Throughout this paper, the following notation is adopted: R and C denote the fields of real and complex numbers, respectively; Rn denotes the n-dimensional real Euclidean space; Rn×n denotes the space of n × n matrices with real entries; xi denotes the ith component of the vector x in Rn ; aij denotes the entry in the (i, j) position of the matrix A in Rn×n . The following notation is also adopted: (i) The matrix A is said to be Hurwitz if all its eigenvalues lie in the open left half of the complex plane. The determinant of the matrix A is denoted det[A]. (ii) A real symmetric matrix P is said to be positive definite if all its eigenvalues are positive. We use P > 0 to denote that P is positive definite. Vectors (matrices) that are entry-wise positive are denoted x  0 (A  0), and vectors (matrices) that are entry-wise nonnegative x  0 (A  0). (iii) We denote the linear time-invariant (LTI) dynamic system associated with the matrix A by ΣA : x˙ = Ax. (iv) Given m linear time-invariant (LTI) dynamic systems, x˙ = Ai x, i ∈ Ω = {1, . . ., m}, the positive definite matrix P is said to be a common Lyapunov solution for Ai if ATi P + PAi = −Qi < 0, i ∈ Ω . In this case V (x) = xT Px defines a common quadratic Lyapunov function (CQLF), for the m LTI systems ΣAi . In the case that P is a diagonal matrix, V (x) is a common diagonal quadratic Lyapunov function (CDLF) for ΣAi . Sometimes, for convenience, but nevertheless abusing notation, we say that a pair of matrices A1 , A2 have a CQLF. By this we mean that the LTI systems ΣA1 , ΣA2 , have a CQLF. The following two results are also directly relevant to the discussions in this paper. (A) A theorem on diagonal stability [2] Redheffer’s result, as given in [2], plays a central role in the developments in this paper. In its simplest form, the result can be paraphrased as follows. Theorem 2.1. Let A ∈ Rn×n be a Hurwitz matrix with negative diagonal entries. Let An−1 denote the [n − 1 × n − 1] leading sub-matrix of A, and Bn−1 denote the corresponding block of A−1 . Then, the matrix A is diagonally stable, if and only if there is a common diagonal Lyapunov function for ΣAn−1 and

ΣBn−1 .

Redheffer’s result was motivated strongly by the observations of Cross a third order systems. The proof of this result is quite involved and is given in [2]. In what follows we provide an alternative proof of this result using the Kalman–Yacubovich–Popov lemma. This proof is shorter, provides greater insights into the diagonal stability problem than the original, and also permits generalisations.

R. Shorten, K.S. Narendra / Linear Algebra and its Applications 431 (2009) 2317–2329

2319

(B) The Meyer version of the KYP lemma The classical Kalman–Yacubovich–Popov (KYP) lemma for single-input systems is found in most text-books on linear dynamic systems. The Meyer version of the lemma [13] needed for the purpose of this note is stated here. Theorem 2.2. Let A ∈ Rn×n be a Hurwitz matrix, and let b, c ∈ Rn . Then there exists a real matrix P P T > 0, a matrix L = LT ∈ Rn×n , L > 0, a vector q ∈ Rn and a scalar γ > 0 such that

+ PA = −qqT − L, √ Pb − c = γ q AT P

(2) (3)

if and only if the rational transfer function H (jω) Real [H (jω)] for all ω

=

= 12 γ + c T (jωI − A)−1 b satisfies

>0

(4)

∈ R [14].

Comment: The KYP lemma plays an important role in systems theory; in particular in the study of the absolute stability problem [15]. It also establishes necessary and sufficient conditions for the existence of a common quadratic Lyapunov function for a pair of linear time invariant systems whose system matrices differ by a matrix of rank one. Indeed, one can show that Eq. (4) is necessary and sufficient for the dynamic systems x˙ = Ax,  x˙ =

A−

2

γ

(5)

 bc T

x,

(6)

to have a common quadratic Lyapunov function. This observation lies at the heart of the circle criterion [16–18]. Thus, one may view the KYP lemma as giving conditions under which a pair of dynamic systems, whose system matrices differ by a rank one matrix, have a common quadratic Lyapunov function. This resembles the statement in Theorem 2.1 (Redheffer) in which conditions were specified under which a pair of dynamic systems (ΣAn−1 and ΣBn−1 )have a common diagonal quadratic Lyapunov function. In fact, since any quadratic Lyapunov function for ΣA is a quadratic Lyapunov function for ΣA−1 [19], it immediately follows that the conditions are also necessary and sufficient for (ΣAn−1 and ΣB−1 ) to n−1

have a common diagonal quadratic Lyapunov function. Thus, one obtains the following corollary to the Theorem of Redheffer. Corollary 2.1. Let A ∈ Rn×n be a Hurwitz matrix with negative diagonal entries. Let An−1 denote the [n − 1 × n − 1] leading sub-matrix of A, and Bn−1 denote the corresponding block of A−1 . Then, the matrix A is diagonally stable, if and only if there is a common diagonal Lyapunov function for ΣAn−1 and

ΣB−1 . n−1

Finally, it will be useful to use a version of the KYP lemma based on Linear Matrix Inequalities [20,21], which we simply state here. More details on this version of the KYP lemma can be found in the afore-mentioned references and in many other recent texts on control theory. LMI version of KYP lemma: Let A ∈ Rn×n be a Hurwitz matrix. Let B ∈ Rn×m , C T ∈ Rm×n and D ∈ Rm×m . Further, let D + DT > 0. Then, there exists a positive definite matrix P = P T ∈ Rn×n , P > 0 such that



A −C T

B −D

T 

P 0



0 I



P

+ 0



0 I

A −C T

if and only if H (jω) + H (jω)∗ > 0 with H (jω)

B −D



< 0.

= C T (jω − A)−1 B + D for all ω ∈ R.

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So if the Hermitian matrix H (jω) + H (jω)∗ is positive definite for all frequencies, then the matrix   A B −C T −D is Hurwitz and has a block diagonal Lyapunov function   P 0  P= . 0 I This version of the lemma will be the key in rederiving and extending the Redheffer theorem on diagonal stability. 3. Principal result Before we state the main result of our paper we note the following result [22] on matrix inverses that plays an important role in the remainder of our paper. Given, A ∈ Rn×n , an invertible real matrix that is partitioned as   A n −1 b n −1 , (7) A= T c n −1 d n −1 where An−1 ∈ R(n−1)×(n−1) , bn−1 , cn−1 ∈ Rn−1 , dn−1 ∈ R and the subscript (n − 1) denotes the dimensions of the matrix An−1 and the vectors bn−1 , cn−1 , respectively. Further assume that dn−1 = / 0. Then, A−1 can be expressed in partitioned form as   B n −1 l n −1 , (8) A −1 = mTn−1 ηn−1 where Bn−1 Bn−1

∈ R(n−1)×(n−1) , ln−1 , mn−1 ∈ Rn−1 and ηn−1 ∈ R and where 

= A n −1 −

bn−1 cnT−1

 −1

.

dn −1

This latter fact is well known and can be easily seen as follows. Since AA−1 the unit matrix in Rn×n , it follows that A n −1 B n − 1

+ bn−1 mTn−1 = In−1

¯ n −1 l n − 1 + η n −1 b n −1 = 0 A cnT−1 Bn−1 + dn−1 mTn−1 = 0 cnT−1 ln−1

+ dn−1 ηn−1 = 1

(9)

= In , where In denotes (10) (11) (12) (13)

From Eq. (12), we have mTn−1

=−

cnT−1 Bn−1

and it follows from Eq. (10) that   bn−1 cnT−1 A n −1 − B n− 1 d n− 1 or equivalently that  Bn−1

(14)

d n −1

= A n −1 −

= I n− 1

bn−1 cnT−1 dn −1

 −1 (15)

as we have claimed. In particular, the form of Bn−1 in the previous lemma will play an important role in the statements of our main result, which we now state as Theorem 3.1.

R. Shorten, K.S. Narendra / Linear Algebra and its Applications 431 (2009) 2317–2329

Theorem 3.1. Let A be a matrix partitioned as   A n −1 b n −1 , A= T c n −1 d n −1 where An−1 En−1

2321

(16)

∈ R(n−1)×(n−1) , bn−1 , cn−1 ∈ Rn−1 , dn−1 ∈ R, dn−1 < 0 and 

= A n −1 −

bn−1 cnT−1



.

d n −1

(17)

+ Dn−1 An−1 < 0 and (En−1 )T Dn−1 + Dn−1 En−1 < 0 for some positive definite diagonal matrix Dn−1 ∈ Rn−1×n−1 ; i.e. if both An−1 and En−1 have a common diagonal Lyapunov function (CDLF).

Then, A is diagonally stable if and only if ATn−1 Dn−1

Proof. (a) Necessity: Let A be a diagonally stable matrix. Then so is A−1 . Furthermore, if V (x) = xT Dx is a diagonal Lyapunov function for ΣA , it is also a diagonal Lyapunov function for ΣA−1 where we denote   D n− 1 0 D= 0 α , where Dn−1 is a positive diagonal matrix of dimension [n − 1 × n − 1] and α > 0. It therefore follows from Sylvester’s criterion that ATn−1 Dn−1 + Dn−1 An−1 < 0, and that BnT −1 Dn−1 + Dn−1 Bn−1 < 0 where Bn−1 was defined in the preamble to the statement of the main theorem. But if Bn−1 is diagonally stable −1 then so is Bn−1 with the same Lyapunov function. Hence, a necessary condition for the matrix A to be diagonally stable is that ΣAn−1 and ΣB−1 have a common diagonal Lyapunov function. Since A is n−1

−1

diagonally stable (and hence invertible), we have that En−1 = Bn−1 . This completes the necessity part of the proof. (b) Sufficiency: Let there exist a diagonal matrix Dn−1 > 0 that simultaneously satisfies ATn−1 Dn−1  A n −1

−

+ D n −1 A n −1 < 0 ; bn−1 cnT−1 dn −1

(18)

T

D n −1



+ D n −1 A n −1 −

bn−1 cnT−1



d n −1

< 0.

(19)

To establish sufficiency we wish to show that there exists a positive scalar α > 0 such that  T      An−1 b n −1 b n −1 D n −1 D n −1 0 0 A n −1 + T T 0 0 α α cn−1 dn−1 < 0, c n −1 d n −1 or equivalently  Dn−1 An−1 cnT−1

D n −1 b n −1 d n −1

T 

I 0

0





I

α + 0

0



α

D n −1 A n −1 cnT−1

D n − 1 b n −1 d n −1



< 0.

It follows from the Kalman–Yacubovich–Popov lemma, that the above matrix inequality is satisfied if and only if the matrix 



− ATn−1 Dn−1 + Dn−1 An−1 +

Dn−1 bn−1 cnT−1 j ω + d n −1

+

cn−1 bTn−1 Dn−1

−jω + dn−1

>0

(20)

(is positive definite for all frequencies). Eqs. (18) and (19) imply that the matrix defined in Eq. (20) has positive eigenvalues for ω = 0 and for ω = ±∞. Now let there exist ω = ωc such that Eq. (21) ceases to be positive definite. This means that at least one eigenvalue of 



− ATn−1 Dn−1 + Dn−1 An−1 +

Dn−1 bn−1 cnT−1 jωc

+ d n −1

+

cn−1 bTn−1 Dn−1

−jωc + dn−1

(21)

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is non-positive. However, since the eigenvalues of a Hermitian matrix are always real, it follows by continuity that the matrix (21) will have a negative eigenvalue at ωc if and only if

  Dn−1 bn−1 cnT−1 cn−1 bTn−1 Dn−1 =0 (22) + det − ATn−1 Dn−1 + Dn−1 An−1 + j ω + d n −1 −jω + dn−1 for some real ω = ω∗ . Assuming that such an ω = ω∗ exists, it follows that

  Dn−1 bn−1 cnT−1 cn−1 bTn−1 Dn−1 T + det − An−1 Dn−1 + Dn−1 An−1 + j ω ∗ + d n −1 −jω∗ + dn−1 Then,

det [M] det I

+

M −1 Dn−1 bn−1 cnT−1 

jω ∗

+ d n −1

+

M −1 cn−1 bTn−1 Dn−1

= 0.



−jω∗ + dn−1

= 0,

(23)



= − ATn−1 Dn−1 + Dn−1 An−1 is invertible since it is positive definite by assumption. Consequently, det [M] × where the matrix M ⎡

det ⎣I

+

M −1 D

n −1 b n −1

, M −1 c

 n −1

This latter equation implies det [M] × ⎡ ⎡ 1 0 ∗ det ⎣I2×2 + ⎣ jω +dn−1 1 0 −j ω ∗ +d

⎡

1

∗ ⎣ jω +dn−1 0

0 1

−jω∗ +dn−1

⎤ ⎦

⎤ cnT−1 ⎦ bTn−1 Dn−1

= 0.

⎤ n−1

where we have used the fact that det[In×n

⎤

 T c − 1 − 1 − 1 n ⎦ M D n −1 b n −1 , M c n −1 ⎦ bTn−1 Dn−1

= 0,

+ UV ] = det[Ip×p + VU ], with U ∈ Rn×p and V ∈ Rp×n .

 −1  2 × Finally, this latter equation can be written as det [M] ω∗ + dn2−1

cnT−1 M −1 cn−1 jω∗ + dn−1 + cnT−1 M −1 Dn−1 bn−1 det bTn−1 Dn−1 M −1 Dn−1 bn−1 −jω∗ + dn−1 + bTn−1 Dn−1 M −1 cn−1

= 0.

Since det[M ] = / 0 (M is positive definite) and ω∗ + dn2−1 > 0 (for all ω), the above equation can only be satisfied if

cnT−1 M −1 cn−1 dn−1 + cnT−1 M −1 Dn−1 bn−1 det bTn−1 Dn−1 M −1 Dn−1 bn−1 dn−1 + bTn−1 Dn−1 M −1 cn−1 2

is a negative number. But this determinant of the matrix defined by (22) evaluated at ω = 0, and scaled by dn2−1 /det[M ], which is positive by assumption. Hence, ω∗ does not exist and by contradiction sufficiency is proven; namely that if An−1 and En−1 share a common quadratic Lyapunov function then the matrix A is diagonally stable.  Comment: The following two points are worth making concerning the principal result given in Theorem 3.1. (i) Redheffer demonstrated that if An−1 and Bn−1 share a CDLF, and if dn−1 < 0, then A is diagonally stable. In Theorem 3.1, necessary and sufficient conditions for Diagonal Stability are given in −1 −1 terms of An−1 and Bn−1 (rather than Bn−1 ). Recall that Bn−1 is a matrix that differs from An−1 by a matrix of rank 1. Further, this matrix is constructed directly from bn−1 , cn−1 and dn−1 . Consequently, Theorem 3.1, provides insight as to the nature of diagonal stability of A directly in terms its elements. It should also be noted that the question as to whether a Lyapunov function exists for a pair of dynamic systems, whose system matrices differ by a matrix that is of unit rank, also arises in the absolute stability problem [15] and has been extensively investigated. Some of the results derived in this context are also applicable in the context of diagonal stability.

R. Shorten, K.S. Narendra / Linear Algebra and its Applications 431 (2009) 2317–2329

2323

(ii) A necessary condition for the Hurwitz matrix A to be diagonally stable is that An−1 is diagonally stable. However, it is seen that the diagonal stability of An−1 is not explicitly made use of in the sufficiency part of the proof; rather the key point that we exploit is the existence of a Lyapunov function (which happens in the present case to be diagonal). Consequently, one may ask whether the proof extends to more general situations. Specifically, given a matrix A, whether necessary and sufficient conditions can be derived, such that there exits a P = P T > 0 satisfying AT P + PA < 0, with   P 0 P = n −1 , 0 1 and where Pn−1 is constrained in some specific manner. By employing the method of proof given in Theorem 3.1, one can show that the necessary and sufficient condition for such a P to exist is that Pn−1 constrained in the aforementioned manner satisfies ATn−1 Pn−1

+ Pn−1 An−1 < 0,

BnT −1 Pn−1

+ P n − 1 B n −1 < 0 .

4. Applications of the main result The result of Redheffer [2], and the principal result of this paper given in Theorem 3.1, both replace the problem of determining a diagonal Lyapunov function for a matrix A to the equivalent problem of determining a common diagonal Lyapunov function (CDLF) for two lower dimensional matrices. In general, the latter problem is not simpler than the former. However, in special cases, due to the −1 structure of the two matrices An−1 and Bn−1 described in Section 3, the determination of a CDLF is substantially simplified. We treat some of these special classes of matrices in this section. Metzler and related matrices: A matrix A ∈ Rn×n is called a Metzler matrix if its off diagonal elements are non-negative. Furthermore, a necessary condition for a matrix A to be Hurwitz and Metzler, it that its diagonal entries are negative. More pertinent to the present paper, a Metzler matrix that is Hurwitz stable is also diagonally stable. We now note the following result concerning Metzler matrices. Lemma 4.1. Let A, A + bc T be a pair of Metzler matrices, where b, c are non-negative vectors. Let T    A + bc T D + D A + bc T < 0 for some diagonal matrix D > 0. Then AT D + DA < 0. T    Proof. The proof of the lemma follows immediately from the fact that A + bc T D + D A + bc T   T   AT D + DA. Since both matrices are Metzler, and since A + bc T D + D A + bc T  0, the assertion of the lemma follows immediately from basic properties of Metzler matrices.



A consequence of the previous lemma and our main result is the following theorem. Theorem 4.1. Let the matrix A ∈ Rn×n be a Metzler matrix. Define the sequence of matrices {A[n], A[n − 1], . . .., A[1]} as follows. A[n] = A. For k = 2, . . ., n partition

A k −1 b k −1 A[k] = , (24) ckT−1 dk −1 where Ak−1 A k −1

−

∈ R(k−1)×(k−1) , bk−1 , ck−1 ∈ Rk−1 . The matrix A[k − 1] is defined to be A[k − 1] =

bk−1 ckT−1 . dk −1

Then, a necessary and sufficient condition for the matrix A to be Hurwitz, is that the

diagonal entries of the matrices A[1], .., A[n] are all strictly negative.

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Proof. The assertions of the Theorem follow from Theorem 3.1, Lemma 4.1, and from the fact that a Metzler matrix is Hurwitz if and only if it is diagonally stable [23]. If A is diagonally stable, by the principal theorem, the dynamic systems   bn−1 cnT−1 x˙ = An−1 x, and x˙ = An−1 − x d n −1 must have a common diagonal Lyapunov function. But, since A[n − 1] is Metzler, and since bn−1 , cn−1 are both entrywise nonnegative, by Lemma 4.1 these dynamic systems will have a common diagonal Lyapunov function if and only if A[n − 1] is diagonally stable. Thus, A is Hurwitz if and only if A[n − 1] is diagonally stable (or equivalently Hurwitz). But by the previous argument A[n − 1] is Hurwitz stable if and only if A[n − 2] is diagonally stable. Hence, it follows that the diagonal entries of A[n − 2] must also be negative. By repeating this argument for all the A[i], i = n, . . ., 2 one concludes that the diagonal entries of all of the aforementioned matrices must be negative and this constitutes the necessary part of the proof. To observe the sufficiency part of the proof assume that all the A[k] have negative diagonal entries. Consider now A[1] and A[2]. Since A[1] is negative, and since A[2] has negative diagonal entries, then this implies that A[2] is Hurwitz. To see this note that A[1] is a scalar, and the (1, 1) entry of A[2] is also a negative scalar. Thus, they have a common diagonal Lyapunov functions and this in turn implies that A[2] is Hurwitz (from the matrix version of the KYP lemma). By repeating this argument for A[3], A[4] and so on, one concludes that the condition given in the statement of the main theorem is also sufficient for stability.  A nice feature of the proof of the previous theorem is that it provides a constructive procedure to check whether a Metzler matrix is Hurwitz. Namely, given a Metzler matrix, we recursively construct its Schur complement. At every stage we only verify whether or not the diagonal elements of the matrix of lower dimension are negative. The process is continued untill only a single element remains. A necessary and sufficient condition for A to be diagonally stable is that this element is negative. The following example illustrates the procedure discussed above. Example 4.1. Let A be a Metzler matrix in R4×4 ⎡ ⎤ −3 1 2 1 ⎢ 1 −6 1 2 ⎥ ⎥. A[4] = ⎢ ⎣ 3 2 −6 1 ⎦ 2 1 3 −10 Then

⎡

−3

A3

=⎣ 1

3 and d3

1 −6 2

⎤ 2 1 ⎦ b3 −6

= −10,

A[3]

= A3 −

b3 c3T d3

⎡

=

=

−2.8 ⎣ 1.4 3.2

⎡ ⎤ 1 ⎣2⎦ c3 1 1.1 −5.8 2.1

=

(25)

⎡ ⎤ 2 ⎣1⎦ 3

⎤ 2.3 1.6 ⎦ −5.7

(26)

(27)

To determine whether A[3] is diagonally stable we repeat the operation to reduce it to a matrix in R2×2 Let       2.3 3.2 −2.8 1.1 A2 = , b2 = , c2 = (28) 1.6 2.1 1.4 −5.8

= −5.7. Then  −1.51 A[2] = 2.30

and d2



1.95 −5.21

(29)

A[2] is Hurwitz, Metzler and hence diagonally stable. This in turn ensures that A is diagonally stable. Example 4.2. If in the previous example the element a44 is chosen to be 5, it follows that

R. Shorten, K.S. Narendra / Linear Algebra and its Applications 431 (2009) 2317–2329

⎡

−2.6 A[3] = ⎣ 1.8 3.4

1.2 −5.6 2.2

which results in  −0.963 A[2] = 3.185

⎤ 2.6 2.2 ⎦ −5.4

2.25 −4.704

2325

(30)

 (31)

which is not Hurwitz. This implies that A is not diagonally stable. Comment: We note briefly that the diagonal stability test given in Theorem 4.1 (with minor variations) can also be applied to other matrix classes. An example for which our Theorem is useful are matrices for which An−1 has the sign pattern of a Metzler matrix, dn−1 < 0 and for which bn−1 and cn−1 are strictly entry-wise non-positive. Thus, the diagonal stability of A can be deduced by examining the signs of the diagonal entries of the matrix sequence {A[n], A[n − 1], . . .., A[1]}. In fact this test for diagonal stability can be applied to any matrix of the form A = Θ M Θ where Θ is a diagonal matrix with diagonal entries that are non-zero, and where M is Metzler. Here the key point to note is that the sign of the diagonal entries of A[n − 1] are negative if and only if the diagonal entries of the corresponding matrix M [n − 1] are as-well. Comment (Symmetric matrices): Any symmetric matrix A that is Hurwitz stable, is also diagonally stable, with a Lyapunov function V (x) = xT x (D = I the unit matrix). From Theorem 3.1 the matrix A is

−

diagonally stable if and only if An−1 , and An−1

bn−1 cnT−1 dn−1

have a common diagonal Lyapunov function.

is a Lyapunov function of the latter, it is also Since cn−1 = bn−1 , it readily follows that if a Lyapunov function of the former. Or alternatively, a symmetric matrix A is Hurwitz stable, if and xnT −1 xn−1

only if An−1

−

bn−1 cnT−1 dn−1

is Hurwitz. This process of reducing the dimensionality of the matrix may be

continued until either one or more diagonal elements of the matrix at every stage is positive, or the final scalar obtained is negative. In the former case the matrix is unstable, while in the latter case, the matrix is stable. This result can also be deduced from the outer product form of the Cholesky decomposition of a symmetric matrix [24]. However, it is of interest to note that it follows as an immediate consequence of Theorem 3.1. Example 4.3. If the symmetric matrix A[4] is defined as ⎡ ⎤ −6 2 −1 2 ⎢ 2 −4 −2 1 ⎥ ⎥ A[4] = ⎢ ⎣−1 −2 −6 −2⎦ . 2 1 −2 −1

(32)

We successively form the following symmetric matrices in R3×3 , R2×2 , as ⎡ ⎡ ⎤ ⎡ ⎤ ⎤ 4 2 −6 2 −1 −4 −2 4 −5 1 −4 −2⎦ + ⎣ 2 −2⎦ = ⎣ 4 −3 −4⎦ A[3] = ⎣ 2 −1 −2 −6 −4 −2 −5 −4 −2 4 and A[2]

=



−2 4



4 −3

+

 1 25 2 20

20 16



=



10.5 14

 14 , 5

which has positive diagonal elements and hence is not diagonally stable. Cyclic matrices: As stated earlier, the main result re-states the diagonal stability problem for a matrix A, with an equivalent CDLF existence problem for a pair of lower dimensional systems. In general, this latter problem is very difficult (and still open) and the examples presented earlier all exploit special matrix properties to avoid solving this problem explicitly. In some cases however, it is possible to solve the general CDLF problem, and one such problem is presented here.

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R. Shorten, K.S. Narendra / Linear Algebra and its Applications 431 (2009) 2317–2329

In a recent paper [25] a class of systems with a cyclic structure was explored and conditions were stated under which cyclic systems would be diagonally stable. This system class is important as it arises in practice, and because the authors demonstrate that many of their results extend to classes of non-linear dynamic systems with certain interconnection structures. One of the main results given in [25] can be summarised as follows. The secant condition: Let An ∈ Rn×n be the matrix ⎡ −1 1 0 ··· ··· ··· ⎢ 0 − 1 1 0 ··· ··· ⎢ ⎢ 0 0 −1 1 ··· ··· ⎢ An = ⎢ .. .. .. .. .. ⎢ .. ⎢ . . . . . . ⎢ ⎣ 0 0 0 0 · · · −1 −K 0 0 0 ··· 0 where K > 0 i

0 0 0

⎤

⎥ ⎥ ⎥ ⎥ .. ⎥ ⎥, . ⎥ ⎥ 1 ⎦ −1

= 1, 2, . . ., n. The matrix is diagonally stable if and only if

K < (sec(π/n))n . We shall now illustrate how to apply Theorem 3.1 to obtain this result. Thus, we wish to determine −1 when a pair of LTI systems, with system matrices An−1 , Bn−1 ∈ Rn−1×n−1 , have a CDLF with ⎡ ⎡ ⎤ ⎤ −1 1 ··· ··· −1 1 ··· ··· 0 0 ⎢ 0 ⎢ ⎥ −1 1 ··· −1 1 ··· 0 ⎥ 0 ⎥ ⎢ ⎢ 0 ⎥ ⎢ ⎢ ⎥ ⎥ .. .. ⎢ ⎢ ⎥ ⎥ . . 0 0 0 ··· 0 ⎥ · · · 0 ⎢ 0 ⎢ ⎥ −1 An−1 = ⎢ ⎥ , Bn−1 = ⎢ .. ⎥ .. ⎥ .. .. ⎢ .. ⎢ .. ⎥ .. .. .. .. ⎢ . ⎢ . . . . . . ⎥ . ⎥ . . ⎢ ⎢ ⎥ ⎥ ⎣ 0 ⎣ 0 0 0 0 0 −1 1 ⎦ −1 1 ⎦ 0

0

···

0

−1

−K

0

···

0

−1

and where K > 0. −1 Note that while An−1 is Metzler, Bn−1 is not a Metzler matrix. Rather, it is a Metzler matrix that has been perturbed by a matrix of unit rank. The following lemma is useful when studying pairs of matrices of this form. Lemma 4.2. Let A be a Hurwitz Metzler matrix. Let B = A − bc T be a Hurwitz matrix that is not necessarily Metzler, where c T = [α , 0, 0, . . . , 0], for some non-zero α. Let there exist a strictly positive diagonal matrix D such that AT D + DA = −Q1  0 with Q1 singular and irreducible, and that the trailing principal submatrix of Q1 is positive definite. Suppose further that BT D + DB = −Q2  0 and that there is no other diagonal D > 0 satisfying the strict inequalities (Q1 < 0, Q2 < 0). Then, there exists a diagonal matrix Γ , whose diagonal entries are not all zero, such that det (A + Γ BΓ ) = 0. The following facts are needed to apply the above lemma to solve the CDLF problem. (i) The secant condition is necessary for the diagonal stability of An . This follows from the fact that An has eigenvalues with non-negative real part if K (sec(π/n))n . (ii) An is a triangular matrix when K = 0 and is hence diagonally stable [26]. (iii) If a diagonal D  0 satisfies ATn D + DAn = −Qn  0, then all diagonal entries of D are strictly positive. This follows directly from the structure of Qn and the fact that if D has any diagonal entries that are zero, then this would imply that one of the principal sub-matrices of Qn is −1 indefinite. Note also that An−1 and Bn−1 are Hurwitz stable. T (iv) Let Dn−1 > 0. Suppose that An−1 Dn−1 + Dn−1 An−1 = −Qn−1  0. Then, the leading principal submatrix of Qn−1 is positive definite. Again, this follows directly from the structure of Qn−1 .

R. Shorten, K.S. Narendra / Linear Algebra and its Applications 431 (2009) 2317–2329

2327

(v) Now let K = K ∗ be the value of K for which there exists a diagonal D > 0 such that ATn D + DAn = −Qn  0, but for which there is no positive diagonal D that satisfies the strict inequality. Then Qn is irreducible. This follows from the fact that the graph associated with Qn is strongly connected if D > 0. Items (iii), (iv) and (v) imply that Lemma 4.2 can be applied to

characterise thesituation when −1 ∗ K = K . Thus, there exists a diagonal matrix Γ such that det An−1 + Γ Bn−1 Γ = 0. Let Γ = diag ([γ1 , . . ., γn−1 ]). Then the determinant:  

−1 det An−1 + Γ Bn−1 Γ = (−1)n−1 K ∗ γ1 γn−1 (1 + γ1 γ2 ) · · · (1 + γn−2 γn−1 )     + 1 + γ12 · · · 1 + γn2−1 .

(33)

This determinant can be further simplified by writing v1T = (1, γ1 ), . . ., vnT −1 = (1, γn−1 ), and by not  ing that vi , vj = vi

vj cos θij , where θij denotes the angles between the vectors. Consequently, 

−1 det An−1 + Γ Bn−1 Γ       = (−1)n−1 K ∗ cos (θ12 ) cos (θ23 ) · · · cos (θ12 + θ23 + · · ·) + 1 . 2 2 1 + γ1 · · · 1 + γn−1 The transcendental part of the identity is minimized when all angles are equal, and when nθ = π . Hence, the above determinant is zero only when K ∗ is equal to the secant condition. Thus for all K > K ∗ there cannot be a diagonal Lyapunov function. For K = 0, An is diagonally stable, and at K = K ∗ , we have the marginal situation characterised by Lemma 4.2. By continuity, for all K < K ∗ , the matrix An is diagonally stable. 5. Conclusions Necessary and sufficient conditions for a matrix A to be diagonally stable are derived using the Kalman–Yacubovich–Popov lemma [13]. These conditions are given in terms of the existence of a common diagonal Lyapunov function for a pair of lower dimensional systems. While these conditions are equivalent to those given by Redheffer [2], they provide more insight into the diagonal stability problem, as the derived conditions are stated directly in terms of the elements of the matrix A. The paper concludes by using the main result to derive conditions for Hurwitz stability (for special classes of matrices) which can be tested in a straightforward manner. Acknowledgements This work was supported by an SFI Walton Fellowship and by SFI PI Award 07/IN.1/1901. The first author is grateful to David Malone, Peter Clifford, and especially to Oliver Mason, for many useful discussions relating to the material presented in this paper. Appendix Proof of Lemma 4.2. (i) First of all note that that Q1 is a matrix of rank n − 1 since it is irreducible. Then this also implies that Q2 is of rank n − 1. This follows directly from the interlacing theorems for symmetric matrices [27]. Thus, we can choose vectors x1 , x2 ∈ Rn , such that Q1 x1 = 0, Q2 x2 = 0. Note that x1 can be chosen to be in the positive orthant since A is a Metzler matrix. (ii) The next stage in the proof follows [28,29] directly and is used to show that there can be no diagonal matrix D with

2328

R. Shorten, K.S. Narendra / Linear Algebra and its Applications 431 (2009) 2317–2329

x1T D Ax1 < 0,

(34)

x2T D Bx2

(35)

< 0.

First of all suppose that there is some D satisfying (34). We shall show that by choosing δ1 0 sufficiently small, it is possible to guarantee that AT (D + δ1 D ) + (D + δ1 D )A is negative definite. Firstly, consider the set

Ω1 = {x ∈ Rn : xT x = 1 and xT D Ax  0}. Note that if the set Ω1 was empty, then any positive constant δ1 > 0 would make AT (D + δ1 D ) + (D + δ1 D )A negative definite. Now, assume that Ω1 is non-empty. The function that takes x to xT D Ax is continuous. Thus Ω1 is closed and bounded, hence compact. Furthermore x1 (or any non-zero multiple of x1 ) is not in Ω1 and thus xT DAx is strictly negative on Ω1 . Let M1 be the maximum value of xT D Ax on Ω1 , and let M2 be the maximum value of xT DAx on Ω1 . Then by the final remark in the previous paragraph, M2 < 0. Choose any constant δ1 > 0 such that

δ1 <

|M2 | = C1 +1

M1

and consider the diagonal matrix D + δ1 D  . By separately considering the cases x vectors x of norm 1

∈ Ω1 and x ∈ / Ω1 , x = 1, it follows that for all non-zero

xT (AT (D + δ1 D ) + (D + δ1 D )A)x < 0 provided 0 < δ1 <

|M2 | . M 1 +1

Since the above inequality is unchanged if we scale x by any non-zero

real number, it follows that AT (D + δ1 D ) + (D + δ1 D )A is negative definite. As A is Hurwitz, this implies that the matrix D + δ1 D is positive definite. The same argument can be used to show that there is some C2 > 0 such that xT (BT (D + δ1 D ) + (D + δ1 D )B)x < 0 for all non-zero x, for 0 < δ1 < C2 . So, if we choose δ > 0 less than the minimum of C1 , C2 , we would have a positive definite diagonal matrix D1

= D + δ D ,

which defined a diagonal CQLF for ΣA and ΣB . Clearly this contradicts the assumption of the Lemma. (iii) As there is no diagonal solution to (34), (35) it follows that x1T D Ax1 < 0

⇐⇒ x2T D Bx2 0

(36)

for diagonal D . It follows from this that x1T D Ax1

= 0 ⇐⇒ x2T D Bx2 = 0.

The expressions x1T D Ax1 , x2T D Bx2 , viewed as functions of D , define linear functionals on the space of diagonal matrices in Rn×n . Moreover, we have seen that the null sets of these functionals are identical. So they must be scalar multiples of each other. Furthermore, (36) implies that they are negative multiples of each other. So there is some constant k > 0 such that x1T D Ax1

= −kx2T D Bx2

√

(37)

for all diagonal D . In fact, we can take k = 1 as we may replace x2 with x2 / k if necessary. On expanding out Eq. (37) (with k = 1) and equating coefficients, it follows that

R. Shorten, K.S. Narendra / Linear Algebra and its Applications 431 (2009) 2317–2329

x1

◦ Ax1 = −x2 ◦ Bx2 .

2329

(38)

But we can write x2 = Γ x1 where Γ is a diagonal matrix. This follows from the fact that Q1 is irreducible and hence x1  0. Then, by direct substitution in (38), and by exploiting linearity of the Haadamard product it follows that det(A + Γ BΓ ) as claimed.

=0



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